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If reactor takes 30 day to consume 4 kg ...

If reactor takes 30 day to consume 4 kg of fuel and each fission gives 185 MeV of usable energy , then calculate the power output.

A

`2.75 xx 10^(10)` W

B

`0.012 xx 10^(10)` W

C

`3.5 xx 10^(10)` W

D

` 7.63 xx 10^(10)` W

Text Solution

Verified by Experts

The correct Answer is:
B
235 amu of uranium gives energy of 185 MeV .
`185 / 235 xx 1.6 xx 10^(-13)` J
also, 1 amu =`1.66 xx 10^(-27)` kg
so, energy by 1 amu or `1.66 xx 10^(-27)`1kg of
= `(185 xx 1.6 xx 10^(-13))/ (235)`
Energy released by 4 kg of
`implies` W = `(185 xx 1.6 xx 10^(-13) xx 4) / (1.66 xx 10^(-27)` xx 235)`
= `(3.035 xx 10^(14))` J
Power of reactor = `(3.035 xx 10^(14))/(30 xx 24 xx 60 xx 60)`
= `(0.012 xx 10^(10))` W
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